Question

The figure shows a quadrilateral ABCD where
LABC = ZBCD = 90°. Given that AB = 4 m and
DA = DC = 7 m, find
(1) angle ADC, (ii) the length of BC.​

Answer 1

(i)64.62°

Draw parallel of BC from A which cut DC name the point

now AB=EC

So DE=3 m (as AB=EC=4)

and AD=7 m

So cos(ADC)=3/7

so angle ADC=64.62°

(ii)6.32m

AE=BC

use pythagoras theorem

AE^2+DE^2=AD^2

AE^2+9=49

AE^2=49-9

AE^2=40

AE=6.32m

sorry if my explanation is a bit complicated